Proof:$X_n\overset{a.s}{\rightarrow}X,Y_n\overset{a.s}{\rightarrow}Y \Rightarrow X_n+Y_n\overset{a.s.}{\rightarrow}X+Y$

Question

I have a simple question but i didn't find a full solution to it anywhere.
I would like to know if my demonstration of the following theorem is correct.
Property:
Let $X_n\left ( \omega \right )\overset{a.s.}{\rightarrow}X\left ( \omega \right )$ and $Y_n\left ( \omega \right )\overset{a.s.}{\rightarrow}Y\left ( \omega \right )$ on $(\Omega ;F;\mathbb{P})$ a probability space. So $X_n\left ( \omega \right ) + Y_n\left ( \omega \right )\overset{a.s.}{\rightarrow}X\left ( \omega \right )+Y\left ( \omega \right )$

Proove:
$S_1=\left \{ \omega \in \Omega:X_n(\omega)\overset{p.w.}{\rightarrow} X(\omega) \cap Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \}\subseteq \left \{ \omega \in \Omega:Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \}$. Furthermore $\mathbb{P}( \left \{ \omega \in \Omega \; : \; X_n(\omega) \overset{p.w.}{\rightarrow}X(\omega) \right \})=1 > \mathbb{P}( \left \{ \omega \in \Omega:Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \})=0 $ by def. So: $ \mathbb{P}(S_1)=0$.
$S_2=\left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \cap Y_n(\omega)\overset{p.w.}{\rightarrow} Y(\omega) \right \}\subseteq \left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \right \}$. So for the same reason as before $\mathbb{P}(S_2)=0$
$S_3=\left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \cap Y_n(\omega)\overset{not \; cvge}{\rightarrow} Y(\omega) \right \}\subseteq \left \{ \omega \in \Omega:X_n(\omega)\overset{not \; cvge}{\rightarrow} X(\omega) \right \}$ So for the same reason as before $\mathbb{P}(S_3)=0$

Now by definition the set of all $\omega \in \Omega$ where $X_n(\omega)+Y_n(\omega)\overset{not \; p.w. \; cvge}{\rightarrow}X(\omega)+Y(\omega)$, is include in $S_1\cup S_2\cup S_3$ (as every where $X_n$ and $Y_n$ all ready p.w. converge we have $X_n+Y_n$ p.w. too).
Hence: $\mathbb{P}(\omega \in \Omega : X_n(\omega)+Y_n(\omega)\overset{not \; p.w. \; cvge}{\rightarrow}X(\omega)+Y(\omega))\leq \mathbb{P}(S_1 \cup S_2 \cup S_3)\leq \mathbb{P}(S_1)+\mathbb{P}(S_2)+\mathbb{P}(S_3)=0$
So we just proved that the measure of the subset of $\Omega$ where there is not p.w. convergence of $X_n+Y_n$ has a measure of zero and, hence, the measure of the set where there is p.w. convergence has a size of 1 (by measure property). And this is exactly the defintion of $X_n(\omega)+Y_n(\omega)\overset{a.s.}{\rightarrow} X(\omega)+Y(\omega)$. Q.E.D

Is this correct?
Thank for your help.

Answer 1

The idea is sound, although there is a minor typo towards the end $$\mathbb{P}(X_n+Y_n \text{ does not converge})\le\mathbb{P}(S_1 \cup S_2 \cup S_3)$$ instead of intersections like you have used.

In any case, you've chosen to prove that the set of divergence has measure zero which is fine. There is a slightly different approach that may be a little shorter. If we let $S_1=\{X_n \text{ converges}\}$ and $S_2=\{Y_n \text{ converges}\}$ then $\mathbb{P}(S_1)=\mathbb{P}(S_2)=1$. Thus, $\mathbb{P}(S_1\cap S_2)=1$ and on $S_1\cap S_2$, we know from elementary real analysis that $X_n+Y_n$ converges.